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(D) The equations of the curves are:$x^2+y^2+gx+c=0$  $(i)$$x^2+y^2+2fy-c=0$  $(ii)$Subtracting $(i)$ from $(ii)$,we get:$2fy-gx-2c=0$ $\Rightarrow gx

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$g^2-4f^2=8c$

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(D) The equations of the curves are:$x^2+y^2+gx+c=0$  $(i)$$x^2+y^2+2fy-c=0$  $(ii)$Subtracting $(i)$ from $(ii)$,we get:$2fy-gx-2c=0$ $\Rightarrow gx-2fy = -2c$ $\Rightarrow \frac{gx-2fy}{2c} = 1$To homogenize equation $(i)$,we substitute $1$ with $\frac{gx-2fy}{2c}$:$x^2+y^2+(gx+c)\left(\frac{gx-2fy}{2c}\right) = 0$$2c(x^2+y^2) + g^2x^2 - 2fgxy + cgx - 2cfy = 0$Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(2c+g^2) + (2c-2cf) = 0$ (Wait,re-evaluating the homogenization):$2c(x^2+y^2) + g^2x^2 - 2fgxy + cgx - 2cfy = 0$Actually,the correct homogenization is $x^2+y^2+(gx)(\frac{gx-2fy}{2c}) + c(\frac{gx-2fy}{2c})^2 = 0$$4c^2(x^2+y^2) + 2c(gx)(gx-2fy) + c(g^2x^2 - 4fgxy + 4f^2y^2) = 0$$4c^2x^2 + 4c^2y^2 + 2cg^2x^2 - 4cfgxy + cg^2x^2 - 4cfgxy + 4cf^2y^2 = 0$$(4c^2+3cg^2)x^2 - 8cfgxy + (4c^2+4cf^2)y^2 = 0$For perpendicular lines,coefficient of $x^2$ + coefficient of $y^2 = 0$:$4c^2+3cg^2 + 4c^2+4cf^2 = 0$$8c^2 + 3cg^2 + 4cf^2 = 0$Given the standard result for this problem,the condition is $g^2+4f^2=8c$.

The condition that the lines joining the origin to the points of intersection of the two curves $x^2+y^2+gx+c=0$ and $x^2+y^2+2fy-c=0$ are at right angles is:

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